Question

Match List I with List II.

List I (Conversion)	List II (Number of Faraday required)
A. \(1\) mole of \(\text{H}_2\text{O}\) to \(\text{O}_2\)	I. \(3F\)
B. \(1\) mol of \(\text{MnO}_4^-\) to \(\text{Mn}^{2+}\)	II. \(2F\)
C. \(1.5\) mol of Ca from molten \(\text{CaCl}_2\)	III. \(1F\)
D. \(1\) mol of FeO to \(\text{Fe}_2\text{O}_3\)	IV. \(5F\)

Choose the correct answer from the options given below :

• A. A-II, B-IV, C-I, D-III
• B. A-III, B-IV, C-I, D-II
• C. A-II, B-III, C-I, D-IV
• D. A-III, B-IV, C-II, D-I

Hint:

Find the electrons transferred per mole in each half reaction; one Faraday is one mole of electrons.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
One Faraday (1 F) is the charge of one mole of electrons. So the Faradays needed equals the moles of electrons transferred.

Step 2: Match A:
\(\text{H}_2\text{O} \rightarrow \tfrac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^-\). One mole of water gives 2 electrons, so it needs 2F. A matches II.

Step 3: Match B:
\(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). Manganese goes from +7 to +2, so 5 electrons per mole: 5F. B matches IV.

Step 4: Match C:
\(\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}\). One mole needs 2F, so 1.5 mol needs \(1.5 \times 2 = 3\)F. C matches I.

Step 5: Match D:
FeO to \(\text{Fe}_2\text{O}_3\): \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\), so 1 electron per mole of FeO: 1F. D matches III.

Step 6: Choose the option:
A-II, B-IV, C-I, D-III is option A. The other options mismatch at least one pair, for example B has A-III which would mean 1F for water oxidation.

Final Answer:
The correct matching is A-II, B-IV, C-I, D-III. \[ \boxed{\text{(A) }\text{A-II, B-IV, C-I, D-III}} \]